Metamath Proof Explorer


Theorem pell14qrexpclnn0

Description: Lemma for pell14qrexpcl . (Contributed by Stefan O'Rear, 18-Sep-2014)

Ref Expression
Assertion pell14qrexpclnn0 DAPell14QRDB0ABPell14QRD

Proof

Step Hyp Ref Expression
1 oveq2 a=0Aa=A0
2 1 eleq1d a=0AaPell14QRDA0Pell14QRD
3 2 imbi2d a=0DAPell14QRDAaPell14QRDDAPell14QRDA0Pell14QRD
4 oveq2 a=bAa=Ab
5 4 eleq1d a=bAaPell14QRDAbPell14QRD
6 5 imbi2d a=bDAPell14QRDAaPell14QRDDAPell14QRDAbPell14QRD
7 oveq2 a=b+1Aa=Ab+1
8 7 eleq1d a=b+1AaPell14QRDAb+1Pell14QRD
9 8 imbi2d a=b+1DAPell14QRDAaPell14QRDDAPell14QRDAb+1Pell14QRD
10 oveq2 a=BAa=AB
11 10 eleq1d a=BAaPell14QRDABPell14QRD
12 11 imbi2d a=BDAPell14QRDAaPell14QRDDAPell14QRDABPell14QRD
13 pell14qrre DAPell14QRDA
14 13 recnd DAPell14QRDA
15 14 exp0d DAPell14QRDA0=1
16 pell14qrne0 DAPell14QRDA0
17 14 16 dividd DAPell14QRDAA=1
18 15 17 eqtr4d DAPell14QRDA0=AA
19 pell14qrdivcl DAPell14QRDAPell14QRDAAPell14QRD
20 19 3anidm23 DAPell14QRDAAPell14QRD
21 18 20 eqeltrd DAPell14QRDA0Pell14QRD
22 14 3ad2ant2 b0DAPell14QRDAbPell14QRDA
23 simp1 b0DAPell14QRDAbPell14QRDb0
24 22 23 expp1d b0DAPell14QRDAbPell14QRDAb+1=AbA
25 simp2l b0DAPell14QRDAbPell14QRDD
26 simp3 b0DAPell14QRDAbPell14QRDAbPell14QRD
27 simp2r b0DAPell14QRDAbPell14QRDAPell14QRD
28 pell14qrmulcl DAbPell14QRDAPell14QRDAbAPell14QRD
29 25 26 27 28 syl3anc b0DAPell14QRDAbPell14QRDAbAPell14QRD
30 24 29 eqeltrd b0DAPell14QRDAbPell14QRDAb+1Pell14QRD
31 30 3exp b0DAPell14QRDAbPell14QRDAb+1Pell14QRD
32 31 a2d b0DAPell14QRDAbPell14QRDDAPell14QRDAb+1Pell14QRD
33 3 6 9 12 21 32 nn0ind B0DAPell14QRDABPell14QRD
34 33 expdcom DAPell14QRDB0ABPell14QRD
35 34 3imp DAPell14QRDB0ABPell14QRD